Evaluating (or tightly bounding) $\sum_{j=1}^\infty\frac1{x_j(1+a)^2-a}$, where $x_j$ is the $j$-th positive root of $a\tan\sqrt x=(1+a)\sqrt x$

Question

Say we are given $$\alpha \tan \sqrt x =(1+\alpha) \sqrt x$$ where $\alpha > 0$ ad are after its positive roots. In particular, I am interested in estimating the following $$S=\sum_{j=1}^{\infty}\frac{1}{x_j(1+\alpha)^2-\alpha},$$ where $x_j$ is the j-th positive root.

I know by graphical inspection that the roots are such that $$(j-1)^2\pi^2<x_j<(j-\frac12)^2\pi^2$$ hence I have an upper and a lower bound for the sum where for the summations I use Cauchy stuff. The problem is that the bounds I get are quite bad: numerically, the lower bounds is not good, since, again graphically, I observe that the roots are closer to $(j-\frac12)^2\pi^2$ than to $(j-1)^2\pi^2$ for a wide range of values of $\alpha$.

Here is the question:

Can one find exact values for $S$ in terms of $\alpha$? If not, can we find tight bounds on $S$ (tighter than mine)?

Answer 1

Just for the roots !

Let $x=t^2$ and $k=\frac{1+\alpha}\alpha$ and to remove the discontinuities, search to the zero's of function $$f(t)=k ~t \cos (t)-\sin (t)$$ Expanding around $t= \left(n+\frac{1}{2}\right)\pi$ and using series reversion, $$t_n=\frac{2n+1}{2}\pi-\frac{2}{ k (2 n+1)\pi}-\frac{4 (2 k-1)}{k^3 (2 n+1)^3 \,\pi^3}+\cdots$$